Question Number 127214 by mohammad17 last updated on 27/Dec/20
$${what}\:{is}\:{the}\:{value}\:{of}\:{and}\:{how}\:{can}\:{combute}? \\ $$$$ \\ $$$${erf}\left(\pi\right)? \\ $$
Answered by Olaf last updated on 27/Dec/20
$$\mathrm{erf}\left({x}\right)\:=\:\frac{\mathrm{2}}{\:\sqrt{\pi}}\int_{\mathrm{0}} ^{{x}} {e}^{−{t}^{\mathrm{2}} } {dt} \\ $$$$\left(\mathrm{used}\:\mathrm{in}\:\mathrm{probabilities}\:\mathrm{and}\:\mathrm{statistics}\right) \\ $$$$\mathrm{erf}\left(\pi\right)\:=\:\frac{\mathrm{2}}{\:\sqrt{\pi}}\int_{\mathrm{0}} ^{\pi} {e}^{−{t}^{\mathrm{2}} {dt}} \\ $$$$\mathrm{We}\:\mathrm{can}\:\mathrm{compute}\:: \\ $$$$\mathrm{erf}\left({x}\right)\:=\:\frac{\mathrm{2}}{\:\sqrt{\pi}}\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}} }{\left(\mathrm{2}{n}+\mathrm{1}\right){n}!}{x}^{\mathrm{2}{n}+\mathrm{1}} \\ $$$$\mathrm{erf}\left(\pi\right)\:=\:\frac{\mathrm{2}}{\:\sqrt{\pi}}\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}} }{\left(\mathrm{2}{n}+\mathrm{1}\right){n}!}\pi^{\mathrm{2}{n}+\mathrm{1}} \\ $$$$\mathrm{erf}\left(\pi\right)\:\approx\:\mathrm{2}\sqrt{\pi}−\frac{\mathrm{2}\pi^{\mathrm{5}/\mathrm{2}} }{\mathrm{3}}+\frac{\pi^{\mathrm{9}/\mathrm{2}} }{\mathrm{5}}−\frac{\pi^{\mathrm{13}/\mathrm{2}} }{\mathrm{21}}+{o}\left(\pi^{\mathrm{17}/\mathrm{2}} \right) \\ $$
Commented by mohammad17 last updated on 27/Dec/20
$${thank}\:{you}\:{sir} \\ $$